# RS Aggarwal Class 10 Solutions Chapter 15 - Probability Ex 15A (15.1)

## RS Aggarwal Class 10 Chapter 15 - Probability Ex 15A (15.1) Solutions Free PDF

The RS Aggarwal solutions of Chapter 15 Ex 15A help students in understanding the basic concepts of mathematics with ease. The solutions provide detailed step by step explanations of each question mentioned in the RS Aggarwal Maths textbook. It helps in building a strong foundation when learning difficult maths concepts. Having proper knowledge of the concepts will help you to score better marks in your maths exam.

Students of Class 10 are advised to solve the RS Aggarwal Class 10 solutions as it covers the entire syllabus as per the CBSE. Students can have a quick revision and analyze their preparation level. Solving the solutions will help you to improve your performance.

## Download PDF of RS Aggarwal Class 10 Solutions Chapter 15â€“ Probability Ex 15A (15.1)

QUESTION 1: Two different dice are rolled simultaneously. Find the probability that the sum of the numbers on the two dice is 10.

Solution:

When two different dice are thrown, the total number of outcomes = 36.

Let E1 be the event of getting the sum of the numbers on the two dice is 10.

These numbers are (4, 6), (5, 5) and (6, 4).

Number of favorable outcomes = 3

Therefore, P(getting the sum of the numbers on the two dice is 10) = P(E1) = $\frac{Number\, of\, outcomes\, favorable\, to\, E_{1}}{Number\, of\, all\, possible\, outcomes}$ = $\frac{3}{36}$ = $\frac{1}{12}$

Thus, the probability of getting the sum of the numbers on the two dice is 10 is $\frac{1}{12}$.

QUESTION 2:When two dice are tossed together, find the probability that the sum of the numbers on their tops is less than 7.

Solution:

When two different dice are thrown, the total number of outcomes = 36.

Let E be the event of getting the sum of the numbers less than 7.

These numbers are (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (2, 1), (2, 2), (2, 3), (2, 4), (3, 1), (3, 2), (3, 3), (4, 1), (4, 2) and (5, 1).

Number of favorable outcomes = 15

Therefore, Â P(getting the sum of the numbers less than 7) = P(E) = $\frac{Number\, of\, outcomes\, favorable\, to\, E}{Number\, of\, all\, possible\, outcomes}$ = $\frac{15}{36}$ = $\frac{5}{12}$

Thus, the probability of getting the sum of the numbers less than 7 is $\frac{5}{12}$.

QUESTION 3:Two dice are rolled together. Find the probability of getting such numbers on two dice whose product is perfect square.

Solution:

When two different dice are thrown, then total number of outcomes = 36.

Let E be the event of getting the product of numbers, as a perfect square.

These numbers are (1, 1), (1, 4), (2, 2), (3, 3), (4, 1), (4, 4), (5, 5) and (6, 6).

Number of favorable outcomes = 8

Therefore, P(getting the product of numbers, as a perfect square) = P(E) = $\frac{Number\, of\, outcomes\, favorable\, to\, E}{Number\, of\, all\, possible\, outcomes}$ = $\frac{8}{36}$ = $\frac{2}{9}$

Thus, the probability of getting the product of numbers, as a perfect square is $\frac{2}{9}$.

QUESTION 4: Two dice are rolled together. Find the probability of getting such numbers on the two dice whose product is 12.

Solution:

Number of all possible outcomes = 36

Let E be the event of getting all those numbers whose product is 12.

These numbers are (2, 6), (3, 4), (4, 3) and (6, 2).

Therefore, P(getting all those numbers whose product is 12) = P(E) = $\frac{Number\, of\, outcomes\, favorable\, to\, E}{Number\, of\, all\, possible\, outcomes}$ = $\frac{4}{36}$ = $\frac{1}{9}$

Thus, the probability of getting all those numbers whose product is 12 is $\frac{1}{9}$.

QUESTION 5:Cards marked with numbers 5 to 50 are placed in a box and mixed thoroughly. A card is drawn from the box at random. Find the probability that the number on the taken out card is

(i) a prime number less than 10

(ii) a number which is a perfect square.

Solution:

All possible outcomes are 5, 6, 7, 8 â€¦â€¦â€¦â€¦â€¦ 50.

Number of all possible outcomes = 46

(i) Out of the given numbers, the prime numbers less than 10 are 5 and 7.

Let E1 be the event of getting a prime number less than 10.

Then, number of favorable outcomes = 2

Therefore, Â P(getting a prime number less than 10) = P(E) = $\frac{2}{46}$ = $\frac{1}{23}$

(ii) Out of the given numbers, the perfect squares are 9, 16, 25, 36 and 49.

Let E2 be the event of getting a perfect square.

Then, number of favorable outcomes = 5

Therefore, P(getting a perfect square) = P(E) = $\frac{5}{46}$

QUESTION 6: A game of chance consists of spinning and arrow which is equally likely to come to the rest pointing to one of the numbers 1, 2, 3, 4, â€¦â€¦, 12 as shown in the figure. What is the probability that it will point to

(i) 6

(ii) An even number

(iii) A prime number

(iv) A number which is a multiple of 5

Solution:

The possible outcomes are 1, 2, 3, 4, 5 â€¦â€¦â€¦â€¦â€¦â€¦.. 12.

Number of all possible outcomes = 12

(i) Let E1 be the event that the pointer rests on 6.

Then, number of favorable outcomes = 1

Therefore, P(arrow pointing at 6) = P(E1) = $\frac{1}{12}$

(ii) Out of the given numbers, the even numbers are

2, 4, 6, 8, 10 and 12

Let E2 be the event of getting an even number.

Then, number of favorable outcomes = 6

Therefore, P(arrow pointing at an even number) = P(E2) = $\frac{6}{12}$ = $\frac{1}{2}$

(iii) Out of the given numbers, the prime numbers are 2, 3, 5, 7 and 11.

Let E3 be the event of the arrow pointing at a prime number.

Then, number of favorable outcomes = 5

Therefore, P(arrow pointing at a prime number) = P(E3) = $\frac{5}{12}$

(iv) Out of the given numbers, the numbers that are multiple of 5 are 5 and 10 only.

Let E4 be the event of the arrow pointing at a multiple of 5.

Then, number of favorable outcomes = 2

Therefore, Â P(arrow pointing at a number that is a multiple of 5) = P(E4) = $\frac{2}{12}$ = $\frac{1}{6}$

QUESTION 7:12 defective pens are accidentally mixed with 132 good ones. It is not possible to just look at pen and tell whether or not it is defective. 1 pen is taken out at random from this lot. Find the probability that the pen taken out is good one.

Solution:

Total number of pens = 132 + 12 = 144

Number of good pens = 132

Let E be the event of getting a good pen.

Therefore, P(getting a good pen) = P(E) = $\frac{Number\, of\, outcomes\, favorable\, to\, E}{Number\, of\, all\, possible\, outcomes}$ = $\frac{132}{144}$ = $\frac{11}{12}$

Thus, the probability of getting a good pen is $\frac{11}{12}$.

QUESTION 8:A lot consists of 144 ballpoint pens of which 20 are defective and others good. Tanvi will buy a pen if it is a good but will not buy if it is defective. The shopkeeper draws 1 pen at random and gives it to her. What is the probability that

(ii) She will not buy it?

Solution:

Total number of pens = 144

Number of defective pens = 20

Number of good pens = 144 â€“ 20 = 124

(i) Let E1 be the event of getting a good pen.

Therefore, P(buying a pen) = P(E1) = $\frac{Number\, of\, outcomes\, favorable\, to\, E_{1}}{Number\, of\, all\, possible\, outcomes}$ = $\frac{124}{144}$ = $\frac{31}{36}$

Thus, the probability that Tanvy will buy a pen is $\frac{31}{36}$.

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(ii) Let E2 be the event of getting a defective pen.

Therefore, P(not buying a pen) = P(E2) = $\frac{Number\, of\, outcomes\, favorable\, to\, E_{2}}{Number\, of\, all\, possible\, outcomes}$ = $\frac{20}{144}$ = $\frac{5}{36}$

Thus, the probability that Tanvy will not buy a pen is $\frac{5}{36}$.

QUESTION 9:A box contains 90 discs which are numbered from 1 to 90 if one disc is drawn at random from the box, find the probability that it bears

(i) A two-digit number

(ii) A perfect square number

(iii) A number divisible by 5.

Solution:

Total number of discs = 90

(i) Let E1 be the event of having a two-digit number.

Number of discs bearing two-digit number = 90 â€“ 9 = 81

Let E1 be the event of getting a good pen.

Therefore, P(getting a two-digit number) = P(E1) = $\frac{Number\, of\, outcomes\, favorable\, to\, E_{1}}{Number\, of\, all\, possible\, outcomes}$ = $\frac{81}{90}$ = $\frac{9}{10}$

Thus, the probability that the disc bears a two-digit number is $\frac{9}{10}$.

(ii) Let E2 be the event of getting a perfect square number.

Disc bearing perfect square numbers are 1, 4, 9, 16, 25, 36, 49, 64 and 81.

Number of discs bearing a perfect square number = 9

Therefore, P(getting a perfect square number) = P(E2) = $\frac{Number\, of\, outcomes\, favorable\, to\, E_{2}}{Number\, of\, all\, possible\, outcomes}$ = $\frac{9}{90}$ = $\frac{1}{10}$

Thus, the probability that the disc bears a perfect square number is $\frac{1}{10}$.

(iii) Let E3 be the event of getting a number divisible by 5.

Discs bearing numbers divisible by 5 are 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85 and 90.

Number of discs bearing a number divisible by 5 = 18

Therefore, P(getting a number divisible by 5) = P(E3) = $\frac{Number\, of\, outcomes\, favorable\, to\, E_{3}}{Number\, of\, all\, possible\, outcomes}$ = $\frac{18}{90}$ = $\frac{1}{5}$

Thus, the probability that the disc bears a number divisible by 5 is $\frac{1}{5}$.

QUESTION 10:(i) A lot of 20 bulbs contain 4 defective ones. 1 bulb is drawn at random from the lot. What is the probability that this bulb is defective?

(ii) Suppose the ball drawn in (i) is not defective and not replaced. Now, ball is drawn at random from the rest. What is the probability that this bulb is not defective?

Solution:

(i) Number of all possible outcomes = 20.

Number of defective bulbs = 4.

Number of non-defective bulbs = 20 â€“ 4 = 16.

Let E1 be the event of getting a defective bulb.

Therefore, Â P(getting a defective bulb) = P(E1) = $\frac{Number\, of\, outcomes\, favorable\, to\, E_{1}}{Number\, of\, all\, possible\, outcomes}$ = $\frac{4}{20}$ = $\frac{1}{5}$

Thus, the probability that the bulb is defective is $\frac{1}{5}$.

(ii) After removing 1 non-defective bulb, we have number of remaining bulbs = 19.

Out of these, number of non-defective bulbs = 16 â€“ 1 = 15.

Let E2 be the event of getting a non-defective bulb.

Therefore, Â P(getting a non-defective bulb) = P(E2) = $\frac{Number\, of\, outcomes\, favorable\, to\, E_{2}}{Number\, of\, all\, possible\, outcomes}$ = $\frac{15}{19}$

Thus, the probability that the bulb is non-defective is $\frac{15}{19}$.

QUESTION 11:A bag contains lemon-flavored candies only. Hema takes out 1 candy without looking into the bag. What is the probability that she takes out

(i) An orange-flavored candy

(ii) A lemon-flavored candy

Solution:

Suppose there are x candies in the bag.

Then, number of orange candies in the bag = 0.

And, number of lemon candies in the bag = x.

(i) Let E1 be the event of getting an orange-flavored candy.

Therefore, P(getting an orange-flavored candy) = P(E1) = $\frac{Number\, of\, outcomes\, favorable\, to\, E_{1}}{Number\, of\, all\, possible\, outcomes}$ = $\frac{0}{x}$ = 0

Thus, the probability that Hema takes out an orange-flavored candy is 0.

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(ii) Let E2 be the event of getting a lemon-flavored candy.

Therefore, P (getting a lemon-flavored candy) = P(E2) = $\frac{Number\, of\, outcomes\, favorable\, to\, E_{2}}{Number\, of\, all\, possible\, outcomes}$ = $\frac{x}{x}$ = 1

Thus, the probability that Hema takes out a lemon-flavored candy is 1.

QUESTION 12:There are 40 students in a class of whom 25 are girls and 15 are boys. The class teacher has to select one student as a class representative. He writes the name of each student on a separate, the card being identical. Then she puts cards in a bag and stirs them thoroughly. She then draws one card from the bag. What is the probability that the name written on the card is the name of

(i) A girl?

(ii) A boy?

Solution:

Total number of students = 40.

Number of boys = 15.

Number of girls = 25.

(i) Let E1 be the event of getting a girlâ€™s name on the card.

Therefore, Â P(selecting the name of a girl) = P(E1) = $\frac{Number\, of\, outcomes\, favorable\, to\, E_{1}}{Number\, of\, all\, possible\, outcomes}$ = $\frac{25}{40}$ = $\frac{5}{8}$

Thus, the probability that the name written on the card is the name of a girl is $\frac{5}{8}$.

(ii) Let E2 be the event of getting a boyâ€™s name on the card.

Therefore, Â P(selecting the name of a boy) = P(E2) = $\frac{Number\, of\, outcomes\, favorable\, to\, E_{2}}{Number\, of\, all\, possible\, outcomes}$ = $\frac{15}{40}$ = $\frac{3}{8}$

Thus, the probability that the name written on the card is the name of a boy is $\frac{3}{8}$.

QUESTION 13:One card is drawn from a well-shuffled deck of 52 cards. Find the probability of drawing

(i) An ace

(ii) A â€˜9â€™ of a black suit

(iv) a red king.

Solution:

Total number of all possible outcomes = 52

(i) Total number of aces = 4

Therefore, P(getting an ace) = $\frac{4}{52}$ = $\frac{1}{13}$

(ii) Number of 4 of spades = 1

Therefore, P(getting a 4 of spade) = $\frac{1}{52}$

(iii) Number of 9 of a black suit = 2

Therefore, P (getting a 9 of a black suit) = $\frac{2}{52}$ = $\frac{1}{26}$

(iv) Number of red kings = 2

Therefore, P (getting a red king) = $\frac{2}{52}$ = $\frac{1}{26}$

QUESTION 14:A card is drawn at random from a well-shuffled deck of 52 cards. Find the probability of getting

(i) A queen

(ii) A diamond

(iii) A king or an ace

(iv) A red ace.

Solution:

Total number of all possible outcomes = 52

(i) Total number of queens = 4

Therefore, P(getting a queen) = $\frac{4}{52}$ = $\frac{1}{13}$

(ii) Number of diamond suits = 13

Therefore, P(getting a diamond) = $\frac{13}{52}$ = $\frac{1}{4}$

(iii) Total number of kings = 4

Total number of aces = 4

Let E be the event of getting a king or an ace card.

Then, the favorable outcomes = 4 + 4 = 8

Therefore, P(getting a king or an ace) = P(E) = $\frac{8}{52}$ = $\frac{2}{13}$

(iv) Number of red aces = 2

Therefore, P(getting a red ace) = $\frac{2}{52}$ = $\frac{1}{26}$

QUESTION 15:One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting

(i) a king of red suit

(ii) a face card

(iii) a red face card

(iv) a queen of black suit

(v) a jack of hearts

Solution:

Total number of outcomes = 52

(i) Let E1 be the event of getting a king of red suit.

Number of favorable outcomes = 2

Therefore, P(getting a king of red suit) = P(E1) = $\frac{Number\, of\, outcomes\, favorable\, to\, E_{1}}{Number\, of\, all\, possible\, outcomes}$ = $\frac{2}{52}$ = $\frac{1}{26}$

Thus, the probability of getting a king of red suit is $\frac{1}{26}$.

(ii) Let E2 be the event of getting a face card.

Number of favorable outcomes = 12

Therefore, P(getting a face card) = P(E2) = $\frac{Number\, of\, outcomes\, favorable\, to\, E_{2}}{Number\, of\, all\, possible\, outcomes}$ = $\frac{12}{52}$ = $\frac{3}{13}$

Thus, the probability of getting a face card is $\frac{3}{13}$.

(iii) Let E3 be the event of getting red face card.

Number of favorable outcomes = 6

Therefore, P(getting a red face card) = P(E3) = $\frac{Number\, of\, outcomes\, favorable\, to\, E_{3}}{Number\, of\, all\, possible\, outcomes}$ = $\frac{6}{52}$ = $\frac{3}{26}$

Thus, the probability of getting a red face card is $\frac{3}{26}$.

(iv) Let E4 be the event of getting a queen of black suit.

Number of favorable outcomes = 2

Therefore, P(getting a queen of black suit) = P(E4) = $\frac{Number\, of\, outcomes\, favorable\, to\, E_{4}}{Number\, of\, all\, possible\, outcomes}$ = $\frac{2}{52}$ = $\frac{1}{26}$

Thus, the probability of getting a queen of black suit suit is $\frac{1}{26}$.

(v) Let E5 be the event of getting a jack of hearts.

Number of favorable outcomes = 1

Therefore, P(getting a jack of hearts) = P(E5) = $\frac{Number\, of\, outcomes\, favorable\, to\, E_{5}}{Number\, of\, all\, possible\, outcomes}$ = $\frac{1}{52}$

Thus, the probability of getting a jack of hearts is $\frac{1}{52}$.

(vi) Let E6 be the event of getting a spade.

Number of favorable outcomes = 13

Therefore, P(getting a spade) = P(E6) = $\frac{Number\, of\, outcomes\, favorable\, to\, E_{6}}{Number\, of\, all\, possible\, outcomes}$ = $\frac{13}{52}$ = $\frac{1}{4}$

Thus, the probability of getting a spade is $\frac{1}{4}$.

### Key Features of RS Aggarwal Class 10 Solutions Chapter 15 â€“Â Probability Ex 15A (15.1)

• It is prepared by subject experts in accordance to the latest CBSE syllabus.
• It is easy to understand and all the solutions are accurately solved for better understanding.
• The RS Aggarwal Maths solutions boost up the student’s confidence level.
• It acts as an important preparation tool and helps students to score good marks in the exam.

#### Practise This Question

Which of the following trends regarding atomic radius is true?